Question

Ahmed and hamad play tennis each week.
The probability that ahmed wins the first match against hamad is 2/3.
What is the probability that ahmed wins exactly three of the next four matches?

Answer 1

Solution:

As, given When Ahmed and Hamad play tennis match against each other, the probability that Ahmed wins the first match against Hamad is $\frac{2}{3}$

Now, we have to find the probability that ahmed wins exactly three of the next four matches.

Let win is denoted by Success and Loss is denoted by failure.

So, Ahmed must get 3 Successes out of 4.

Probability of Success,(S)= $\frac{2}{3}$

Probability of Failure i.e , (F)= 1- $\frac{2}{3}= \frac{1}{3}$

Probability that Ahmed wins exactly three of the next four matches.=

             $=_{3}^{4}\textrm{C}S^3\times F^{4-3}\\\\ =4\times(\frac{2}{3})^3\times(\frac{1}{3})^1 \\\\=4\times\frac{8}{27}\times\frac{1}{3}\\\\ =\frac{32}{81}$

Where, $_{3}^{4}\textrm{C}=\frac{4!}{(4-3)!3!}=\frac{4 \times 3!}{1!\times3!}=4$

Answer 2

Answer:

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